
\prob{002D}{高和中线}

\begin{figure}[htbp]
  \centering
  \image{002D}
  \caption{002D：高和中线} \label{fig:002D}
\end{figure}

如图~\ref{fig:002D}，$OA \perp OA', OB \perp OB'$，$OA = OA', OB = OB'$，$OP \perp AB$于$P$，$Q$是线段$A'B'$的中点，求证：$P, O, Q$三点共线。
\problabels{yellow/平面几何, green/证明题}

\subsection{倍长中线} \label{subsec:002D-mid}

\begin{figure}[htbp]
  \centering
  \image{002D-mid}
  \caption{\nameref{subsec:002D-mid}：通过倍长中线构造两对全等三角形，从而证明。}
  \label{fig:002D-mid}
\end{figure}

如图~\ref{fig:002D-mid}，延长$OQ$到点$O'$，使得$OQ = O'Q$；连接$O'B'$。

\begin{align*}
  \because  {}& \triangle OQA' \cong \triangle O'QB' \ \text{（证明省略）} \\
  \therefore{}& \angle O'B'Q = \angle OA'B', A'O = O'B' \\
  \therefore{}& \angle O'B'Q + \angle OB'A' = \angle OA'B' + \angle OB'A' \\
  \therefore{}& \angle O'B'O = \angle OA'B' + \angle OB'A' \\
  \because  {}& \angle OA'B' + \angle OB'A' + \angle A'OB' = 180^\circ \\
  \therefore{}& \angle O'B'O + \angle A'OB' = 180^\circ \\
  \because  {}& \angle AOB + \angle A'OB' \\
  & + \angle AOA' + \angle BOB' = 360^\circ \\
  \therefore{}& \angle AOB + \angle A'OB' \\
  & = 360^\circ - \angle AOA' - \angle BOB' \\
  \because  {}& OA \perp OA', OB \perp OB' \\
  \therefore{}& \angle AOA' = \angle BOB' = 90^\circ \\
  \therefore{}& \angle AOB + \angle A'OB' = 180^\circ \\
  \therefore{}& \angle AOB = \angle O'B'O \\
  \because  {}& AO = A'O \\
  \therefore{}& AO = O'B' \\
  \because  {}& \text{在 $\triangle AOB$ 与 $\triangle O'B'O$ 中} \\
  & \begin{cases}
    AO = O'B' \\
    \angle AOB = \angle O'B'O \\
    OB = B'O
  \end{cases} \\
  \therefore{}& \triangle AOB \cong \triangle O'B'O \\
  \therefore{}& \angle B'OQ = \angle B \\
  \text{又}\because{}& OP \perp AB \\
  \therefore{}& \angle OPA = 90^\circ \\
  \therefore{}& \angle B + \angle BOP = 90^\circ \\
  \therefore{}& \angle B'OQ + \angle BOP = 90^\circ \\
  \because  {}& \angle POQ = \angle B'OQ + \angle BOP + \angle BOB' \\
  \therefore{}& \angle POQ = 90^\circ + 90^\circ = 180^\circ \\
  \therefore{}& \text{$P, O, Q$ 三点共线}
\end{align*}

证毕。

\subsection{旋转} \label{subsec:002D-rot}

\begin{figure}[htbp]
  \centering
  \image{002D-rot}
  \caption{\nameref{subsec:002D-rot}：通过旋转三角形构造中位线，然后利用位置关系证明。}
  \label{fig:002D-rot}
\end{figure}

基本思路：通过构造一个被旋转的三角形构造一个三角形的中位线，然后利用一系列线段的位置关系证明命题。

如图~\ref{fig:002D-rot}，延长$AO$到点$A''$使得$OA = OA' = OA''$，连接$A''B$；设线段$A''B$的中点为$Q'$，连接$OQ'$。

\begin{align*}
  \because  {}& \angle A'OA'' = \angle BOB' = 90^\circ \\
  \therefore{}& \angle A'OA'' + \angle A''OB' = \angle BOB' + \angle A''OB' \\
  \therefore{}& \angle A'OB' = \angle A''OB \\
  \therefore{}& \triangle A'OB' \cong \triangle A''OB \ \text{（证明省略）} \\
  \therefore{}& A'B' = A''B, \angle B' = \angle OBQ' \\
  \because  {}& B'Q = \frac12A'B', BQ' = \frac12A''B \\
  \therefore{}& B'Q = BQ' \\
  \because  {}& \text{在 $\triangle B'OQ$ 与 $\triangle BOQ'$ 中} \\
  & \begin{cases}
    B'O = BO \\
    \angle B' = \angle OBQ' \\
    B'Q = BQ'
  \end{cases} \\
  \therefore{}& \triangle B'OQ \cong \triangle BOQ' \\
  \therefore{}& \angle B'OQ = \angle BOQ' \\
  \therefore{}& \angle B'OQ + \angle B'OQ' = \angle BOQ' + \angle B'OQ' \\
  \therefore{}& \angle QOQ' = \angle BOB' = 90^\circ \\
  \therefore{}& OQ' \perp OQ \\
  \text{又}\because{}& OA'' = OA, Q'A'' = Q'B \\
  \therefore{}& \text{线段 $OQ'$ 是 $\triangle AA''B$ 的中位线} \\
  \therefore{}& OQ' \parallel AB \\
  \because  {}& OP \perp AB \\
  \therefore{}& OQ' \perp OP \\
  \therefore{}& \text{$P, O, Q$ 三点共线}
\end{align*}

证毕。
